Question

Calculate the pH of a solution made by mixing 49.5 mL of 0.335 M NaA (Ka for HA = 1.0 x 10^-9) with 30.6 mL of 0.120 M HCl.

The answer given was 9.55. But I am stuck on how to get there after the ICE table. Can someone include a step by step.

Answer 1

A- is conjugate base of HA

find Kb for A-

we have below equation to be used:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/1*10^-9

Kb = 1*10^-5

we have:

Molarity of HCl = 0.12 M

Volume of HCl = 30.6 mL

Molarity of A = 0.335 M

Volume of A = 49.5 mL

mol of HCl = Molarity of HCl * Volume of HCl

mol of HCl = 0.12 M * 30.6 mL = 3.672 mmol

mol of A = Molarity of A * Volume of A

mol of A = 0.335 M * 49.5 mL = 16.5825 mmol

We have:

mol of HCl = 3.672 mmol

mol of A = 16.5825 mmol

3.672 mmol of both will react

excess A remaining = 12.9105 mmol

Volume of Solution = 30.6 + 49.5 = 80.1 mL

[A] = 12.9105 mmol/80.1 mL = 0.1612 M

[HA] = 3.672 mmol/80.1 mL = 0.0458 M

They form basic buffer

base is A

conjugate acid is HA

Kb = 1*10^-5

pKb = - log (Kb)

= - log(1*10^-5)

= 5

we have below equation to be used:

This is Henderson–Hasselbalch equation

pOH = pKb + log {[conjugate acid]/[base]}

= 5+ log {4.584*10^-2/0.1612}

= 4.45

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.45

= 9.55

Answer: 9.55